题目描述

N (1 <= N <= 50,000) cows conveniently numbered 1..N are driving in separate cars along a highway in Cowtopia. Cow i can drive in any of M different high lanes (1 <= M <= N) and can travel at a maximum speed of S_i (1 <= S_i <= 1,000,000) km/hour.

After their other bad driving experience, the cows hate collisions and take extraordinary measures to avoid them. On this highway, cow i reduces its speed by D (0 <= D <= 5,000) km/hour for each cow in front of it on the highway (though never below 0 km/hour). Thus, if there are K cows in front of cow i, the cow will travel at a speed of max[S_i - D * K, 0]. While a cow might actually travel faster than a cow directly in front of it, the cows are spaced far enough apart so crashes will not occur once cows slow down as

described,

Cowtopia has a minimum speed law which requires everyone on the highway to travel at a a minimum speed of L (1 <= L <= 1,000,000) km/hour so sometimes some of the cows will be unable to take the highway if they follow the rules above. Write a program that will find the maximum number of cows that can drive on the highway while obeying the minimum speed limit law.

编号为1到N的N只奶牛正各自驾着车打算在牛德比亚的高速公路上飞驰．高速公路有M(1≤M≤N)条车道．奶牛i有一个自己的车速上限Si(l≤Si≤1,000,000)．

在经历过糟糕的驾驶事故之后，奶牛们变得十分小心，避免碰撞的发生．每条车道上，如果某一只奶牛i的前面有南只奶牛驾车行驶，那奶牛i的速度上限就会下降kD个单位，也就是说，她的速度不会超过Si – kD(O≤D≤5000)，当然如果这个数是负的，那她的速度将是0．牛德比亚的高速会路法规定，在高速公路上行驶的车辆时速不得低于/(1≤L≤1,000,000)．那么，请你计算有多少奶牛可以在高速公路上行驶呢？

输入输出格式

输入格式：

• Line 1: Four space-separated integers: N, M, D, and L

• Lines 2..N+1: Line i+1 describes cow i's initial speed with a single integer: S_i

输出格式：

• Line 1: A single integer representing the maximum number of cows that can use the highway

输入输出样例

3 1 1 5

5

7

5

2

说明

There are three cows with one lane to drive on, a speed decrease of 1, and a minimum speed limit of 5.

Two cows are possible, by putting either cow with speed 5 first and the cow with speed 7 second.

 #include<cstdio>

 #include<algorithm>

 #define LL long long

 const int maxn=1e5+;

 LL n,m,ans;

 struct nate{LL t,d;}s[maxn];

 bool comp(nate x,nate y){return x.t*y.d<x.d*y.t;}

 int main(){

     scanf("%lld",&n);

     for(int i=;i<=n;i++){

         scanf("%lld%lld",&s[i].t,&s[i].d);

         s[i].t*=,m+=s[i].d;

     }

     std::sort(s+,s+n+,comp);

     for(int i=;i<=n;i++){

         m-=s[i].d;

         ans+=s[i].t*m;

     }

     printf("%lld\n",ans);

     return ;

 }